Problem: $p(t) = (3\sin(2t), 3\cos(2t), t^2)$ What is the speed of $p(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2\sqrt{9 + t^2}$ (Choice B) B $4\sqrt{t^2}$ (Choice C) C $4\sqrt{25 + 4t^2}$ (Choice D) D $2\sqrt{4 + 2t^4}$
Solution: The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t), c(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 + c'(t)^2 }$ Our position function here is $p(t)$. $\begin{aligned} p'(t) &= (6\cos(2t), -6\sin(2t), 2t) \\ \\ \text{speed} &= ||p'(t)|| \\ \\ &= \sqrt{36\cos^2(2t) + 36\sin^2(2t) + 4t^2} \\ \\ &= \sqrt{36 + 4t^2} \\ \\ &= 2\sqrt{9 + t^2} \end{aligned}$ Therefore, the speed of $p(t)$ is $2\sqrt{9 + t^2}$.